Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e5 + 5;
const long long lg = 17;
const long long mod = 1e9 + 7;
const long long INF = 1e14 + 1;
vector<long long> g[maxn], tt[maxn];
long long dpt[maxn];
vector<long long> up[maxn][lg];
long long par[maxn][lg];
void dfs(long long v, long long p, long long dp) {
  dpt[v] = dp;
  for (long long i = 0; i < min(10ll, (long long)(tt[v]).size()); i++) {
    up[v][0].push_back(tt[v][i]);
  }
  par[v][0] = p;
  for (long long i = 1; i < lg; i++) {
    merge((up[v][i - 1]).begin(), (up[v][i - 1]).end(),
          (up[par[v][i - 1]][i - 1]).begin(), (up[par[v][i - 1]][i - 1]).end(),
          back_inserter(up[v][i]));
    while (up[v][i].size() > 10) up[v][i].pop_back();
    par[v][i] = par[par[v][i - 1]][i - 1];
  }
  for (auto to : g[v]) {
    if (to == p) continue;
    dfs(to, v, dp + 1);
  }
}
vector<long long> vec;
void fuck(long long x, long long i) {
  vector<long long> vec1;
  merge((vec).begin(), (vec).end(), (up[x][i]).begin(), (up[x][i]).end(),
        back_inserter(vec1));
  vec = vec1;
  while ((long long)(vec).size() > 10) vec.pop_back();
}
void lca(long long a, long long b) {
  if (dpt[a] < dpt[b]) swap(a, b);
  for (long long i = lg - 1; i >= 0; i--) {
    if (dpt[a] - (1ll << i) >= dpt[b]) {
      fuck(a, i);
      a = par[a][i];
    }
  }
  if (a == b) {
    fuck(a, 0);
    return;
  }
  for (long long i = lg - 1; i >= 0; i--) {
    if (par[a][i] != par[b][i]) {
      fuck(a, i);
      fuck(b, i);
      a = par[a][i];
      b = par[b][i];
    }
  }
  fuck(a, 1);
  fuck(b, 0);
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m, q;
  cin >> n >> m >> q;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    g[u - 1].push_back(v - 1);
    g[v - 1].push_back(u - 1);
  }
  for (long long i = 0; i < m; i++) {
    long long x;
    cin >> x;
    x--;
    tt[x].push_back(i);
  }
  for (long long i = 0; i < n; i++) sort((tt[i]).begin(), (tt[i]).end());
  dfs(0, 0, 0);
  while (q--) {
    long long v, u, a;
    cin >> v >> u >> a;
    v--, u--;
    lca(u, v);
    sort((vec).begin(), (vec).end());
    vec.erase(unique((vec).begin(), (vec).end()), vec.end());
    while (vec.size() > a) vec.pop_back();
    cout << (long long)(vec).size() << ' ';
    for (auto ii : vec) cout << ii + 1 << ' ';
    cout << '\n';
    vec.clear();
  }
  return 0;
}